[masha_kasha]

I'm doing a practical on finding out the percentage of acetylsalicylic acid in aspirin tablets. I know exactly how to carry out the calculations, only my final answer is over 100% which is obviously wrong. Either my lab results are incorrect, or there's something I'm missing.

I did this using back titration. I poured 25 ml^3 NaOH and the same amount of distilled water onto 1.46 grams of aspirin tablets. After hydrolising (heating it on a burner for 10 minutes), I poured the unused NaOH into a 250 ml^3 flask and made up to mark with distilled water. Then I titrated 25 ml^3 of that solution against HCl of .1 [M] to figure out the volume of that unused NaOH, so I can substract it from the original amount and from an equation find out how much acetylsalicylic acid was used.

My experiment came out to an average of 1.1 ml^3 of HCl. Since the ratio between the two is 1:1, the moles of HCl are .1 x .0011 = .00011, which = the moles of unused NaOH. There was originally .025 moles of NaOH in the solution, therefore .025-.00011=.02489 moles of NaOH were used up in the hydrolysis. From the original equation, the ratio between NaOH and acetylsalicylic acid is 2:1, so that amount of moles / 2 = .012445 moles of the acid. To find the mass, I multiplied that by 180 (molar mass of the acid) and got 2.24. However, the original mass of the aspirin was 1.46! How can I end up with more than i started? What am I doing wrong?

[Yggdrasil]

Remember that your titrations calculate the amount of NaOH in 25mL of your diluted solution.  So, you have 0.00011 mol of NaOH in your 25mL aliquot.  How much NaOH does this correspond to in the total 250mL?

Also, the volume should be mL not mL³.  I think you are getting confused with 1 cm³ which is equal to 1 mL.

[masha_kasha]

Yes, I meant to say ml.

For the aliquots, I don't exactly understand why we have to make it up to mark anyway. But that would be a tenth of the 250ml solution, are you saying I have to multiply .00011 by 10? In that case, .025 - .0011 = .0239. That divided by 2 is .0119, and multiplied by 180 that's still over 1.46 g.

[english]

Yes, I meant to say ml.

For the aliquots, I don't exactly understand why we have to make it up to mark anyway. But that would be a tenth of the 250ml solution, are you saying I have to multiply .00011 by 10? In that case, .025 - .0011 = .0239. That divided by 2 is .0119, and multiplied by 180 that's still over 1.46 g.

Concentration is a constant quantity.  So if you have x mol/L in your aliquot, then you have x mol/L in your 250 mL sample.

[masha_kasha]

Concentration is a constant quantity.  So if you have x mol/L in your aliquot, then you have x mol/L in your 250 mL sample.

Yes, however, the amount of moles does multiply by 10. I have .004 mol/L in my aliquot, and therefore the same amount in the whole solution; .004 x .25 = .001, which is 10 times the moles in the aliquot, which brings me to the same answer of 147%.

[english]

Your number of moles of NaOH must be too high.  0.025 mol sounds way too high.

What was your concentration of NaOH stock from which you took your volumes?  You stated that you added 0.025 L of NaOH, not moles.

[masha_kasha]

Your number of moles of NaOH must be too high.  0.025 mol sounds way too high.

What was your concentration of NaOH stock from which you took your volumes?  You stated that you added 0.025 L of NaOH, not moles.

"Pipette 25cm^3 of the approximately 1.0 mol dm-3 NaOH onto the tablets"

- which equals .025 L x 1 M = .025 moles

[kimi13]

Just a quick clarification, do you know the molarity of NaOH solution you used for titration?

[enahs]

Maybe I do not see it in your calculations and I am just blind, but the very first thing you said you did was dilute your 1M NaOH in half to 0.5M NaOH. You are then titrated a 25mL aliquot of that 0.5M solution. So you first diluted it in half, then you determined the unreacted NaOH of only half of that solution. But I do not see in your calculations where you accounted for that?

[masha_kasha]

Maybe I do not see it in your calculations and I am just blind, but the very first thing you said you did was dilute your 1M NaOH in half to 0.5M NaOH. You are then titrated a 25mL aliquot of that 0.5M solution. So you first diluted it in half, then you determined the unreacted NaOH of only half of that solution. But I do not see in your calculations where you accounted for that?

Are you saying that by adding 25ml of water to the original solution, I diluted the concentration of NaOH in half?

In that case, the original number of moles of added NaOH should have been .5 x .025 instead of 1 x .025... is that what you're saying?
Because that would probably make sense.

[Borek]

That's not correct. Whole reaction mixture was filled up to 250 mL, so it contained 25 mmol of NaOH minus 2 times moles of aspirin.

Do the calculations using number of moles, don't think about dilutions as they will only confuse you.

You start with 25 mmol of NaOH (25mL*1M).

1/10 of the mixture requires 1.1mL of 0.1M HCl to be neutralized. Whole mixture requires 11 mL of 0.1HCl, that is 1.1 mmole of HCl.

25-1.1 = 23.9 mmole of NaOH was used for hydrolysis. There was half of that amount of aspirin present.

That means 11.95 mmole of aspirin - too much (2.15g). I suppose there is some experimental error/data error (wrong concentration). Or we are all missing something obvious.

What indicator have you used to determine endpoint?

[kimi13]

No it was not whole reaction mixture filled up to 250 ml, but as per her description:

I poured the unused NaOH into a 250 ml^3 flask and made up to mark with distilled water. Then I titrated 25 ml^3 of that solution against HCl of .1 [M] to figure out the volume of that unused NaOH

so she used the diluted unused NaOH (in the begining she did diluted  with another 25 ml distilled water, so  now solution has become 0.5M solution), which now refering to the bactitration with 0.1MHCL she took 25 ML of whole flask ,250 ML , which makes 1/10 of all moles presented in flask were titrated.

Now your calculation (Borek):

1/10 of the mixture requires 1.1mL of 0.1M HCl to be neutralized. Whole mixture requires 11 mL of 0.1HCl, that is 1.1 mmole of HCl.



But then : (0.5M  x 25) -1.1 =11.4 mole NaOH was used for hydrolysis. Now we have the ratio 2:1  for the reaction of NaOH with aspirin, which will bring that half of these number is the number of moles of aspirin that took part in hydrolysis, so 11.4/2 = 5.7 mmoles aspirin x 180 Masmolar of aspirin
 = 1.026 g aspirin was hydrolised.

Note: The solution is 0.5M in hydrolysis with aspirin (after adding 25 ML water). Had she not added water the amount of NaOH unused would have been bigger and you would have had different bactitration results.

[Borek]

Perhaps we should ask masha_kasha to clarify things, but IMHO what she (?) means is that the reaction mixture (CONTAINING unused NaOH) was transferred to 250 mL flask.

Otherwise I fail to understand the procedure and its accuracy. You start with 50 mL of 0.5M solution plus some aspirin. Yue heat it for 10 minutes allowing for some unspecified amount of water to leave the mixture. What now - is the procedure to take 25mL of that mixture to the flask? How could you know if it is exactly half, if you don''t know volume after evaporation? OTOH, transferring WHOLE reaction mixture to the 250 mL flask makes perfect sense - volume doesn't matter, moles matter - and they are kept intact.

masha_kasha - please clarify. WHAT volume of WHICH solution was transferred to the 250 mL flask? What do you mean by UNUSED NaOH?

[masha_kasha]

I think I get it.

In the experiment, the ENTIRE SOLUTION after hydrolising (NaOH, water and aspirin) was transferred to the flask. Unused, meaning the NaOH that wasn't used in the hydrolysis of the aspirin. But I think my mistake was that in the very beginning, I calculated .025 L x 1 M = .025 mol NaOH (originally added), but I didn't account for the water. If the same amount of water was added, you guys are saying that that would cut the concentration by 2, therefore it should have been .025 L x .5 M.

That would make much more sense and the percentage works out. Thank you all so much! Especially kimi. Definetely something useful to know for the finals.

[Borek]

masha_kasha, thank you for clarification. Unfortunately, you are wrong. You start with 25 mmoles of NaOH (25mL of 1M solution) and this amount won't change during the experiment. If you dilute it with 25mL of water concentration falls down to 0.5M - but the volume rises up to 50 mL. Doesn't matter if you have 25 mL * 1M, or 50 mL * 0.5M - in both cases you have 25 mmoles of NaOH.

That's simple mass preservation.

Outline the titration procedure. What indicator have you used?